52.3 Formal sections, II

In this section we ask if completion and taking cohomology commute for sheaves of modules on schemes over an affine base $A$ when completion is with respect to a principal ideal in $A$. Of course, we have already discussed the theorem on formal functions in Cohomology of Schemes, Section 30.20. Moreover, we will see in Section 52.7 that derived completion commutes with derived cohomology in great generality. In this section we just collect a few simple special cases of this material that will help us with future developments.

Lemma 52.3.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $f \in \Gamma (X, \mathcal{O}_ X)$. Let

$\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$

be inverse system of $\mathcal{O}_ X$-modules. The following are equivalent

1. for all $n \geq 1$ the map $f : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ to give a short exact sequence $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$,

2. for all $n \geq 1$ the map $f^ n : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_1$ to give a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$

3. there exists an $\mathcal{O}_ X$-module $\mathcal{G}$ which is $f$-divisible such that $\mathcal{F}_ n = \mathcal{G}[f^ n]$.

If $X$ is a scheme and $\mathcal{F}_ n$ is quasi-coherent, then these are also equivalent to

1. there exists an $\mathcal{O}_ X$-module $\mathcal{F}$ which is $f$-torsion free such that $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$.

Proof. We omit the proof of the equivalence of (1) and (2). The condition that $\mathcal{G}$ is $f$-divisible means that $f : \mathcal{G} \to \mathcal{G}$ is surjective. Thus given $\mathcal{F}_ n$ as in (1) we set $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ n$ where the maps $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to \ldots$ are as in (1). This produces an $f$-divisible $\mathcal{O}_ X$-module with $\mathcal{F}_ n = \mathcal{G}[f^ n]$ as can be seen by checking on stalks. The condition that $\mathcal{F}$ is $f$-torsion free means that $f : \mathcal{F} \to \mathcal{F}$ is injective. If $X$ is a scheme and $\mathcal{F}_ n$ is quasi-coherent, then we set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Namely, for an affine open $U \subset X$ the transition maps $\mathcal{F}_{n + 1}(U) \to \mathcal{F}_ n(U)$ are surjective by vanishing of higher cohomology. This produces an $f$-torsion free $\mathcal{O}_ X$-module with $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$ (Lemma 52.2.4). $\square$

Lemma 52.3.2. Suppose $X$, $f$, $(\mathcal{F}_ n)$ is as in Lemma 52.3.1. Then the limit topology on $H^ p = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n)$ is the $f$-adic topology.

Proof. Namely, it is clear that $f^ t H^ p$ maps to zero in $H^ p(X, \mathcal{F}_ t)$. On the other hand, let $c \geq 1$. If $\xi = (\xi _ n) \in H^ p$ is small in the limit topology, then $\xi _ c = 0$, and hence $\xi _ n$ maps to zero in $H^ p(X, \mathcal{F}_ c)$ for $n \geq c$. Consider the inverse system of short exact sequences

$0 \to \mathcal{F}_{n - c} \xrightarrow {f^ c} \mathcal{F}_ n \to \mathcal{F}_ c \to 0$

and the corresponding inverse system of long exact cohomology sequences

$H^{p - 1}(X, \mathcal{F}_ c) \to H^ p(X, \mathcal{F}_{n - c}) \to H^ p(X, \mathcal{F}_ n) \to H^ p(X, \mathcal{F}_ c)$

Since the term $H^{p - 1}(X, \mathcal{F}_ c)$ is independent of $n$ we can choose a compatible sequence of elements $\xi '_ n \in H^1(X, \mathcal{F}_{n - c})$ lifting $\xi _ n$. Setting $\xi ' = (\xi '_ n)$ we see that $\xi = f^{c + 1} \xi '$. This even shows that $f^ c H^ p = \mathop{\mathrm{Ker}}(H^ p \to H^ p(X, \mathcal{F}_ c))$ on the nose. $\square$

Lemma 52.3.3. Let $A$ be a Noetherian ring complete with respect to a principal ideal $(f)$. Let $X$ be a scheme over $\mathop{\mathrm{Spec}}(A)$. Let

$\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$

be an inverse system of $\mathcal{O}_ X$-modules. Assume

1. $\Gamma (X, \mathcal{F}_1)$ is a finite $A$-module,

2. the equivalent conditions of Lemma 52.3.1 hold.

Then

$M = \mathop{\mathrm{lim}}\nolimits \Gamma (X, \mathcal{F}_ n)$

is a finite $A$-module, $f$ is a nonzerodivisor on $M$, and $M/fM$ is the image of $M$ in $\Gamma (X, \mathcal{F}_1)$.

Proof. By Lemma 52.3.2 and its proof we have $M/fM \subset H^0(X, \mathcal{F}_1)$. From (1) and the Noetherian property of $A$ we get that $M/fM$ is a finite $A$-module. Observe that $\bigcap f^ nM = 0$ as $f^ nM$ maps to zero in $H^0(X, \mathcal{F}_ n)$. By Algebra, Lemma 10.96.12 we conclude that $M$ is finite over $A$. $\square$

Lemma 52.3.4. Let $A$ be a ring. Let $f \in A$. Let $X$ be a scheme over $\mathop{\mathrm{Spec}}(A)$. Let

$\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$

be an inverse system of $\mathcal{O}_ X$-modules. Assume

1. either $H^1(X, \mathcal{F}_1)$ is an $A$-module of finite length or $A$ is Noetherian and $H^1(X, \mathcal{F}_1)$ is a finite $A$-module,

2. the equivalent conditions of Lemma 52.3.1 hold.

Then the inverse system $M_ n = \Gamma (X, \mathcal{F}_ n)$ satisfies the Mittag-Leffler condition.

Proof. Set $I = (f)$. We will use the criterion of Lemma 52.2.1. Observe that $f^ n : \mathcal{F}_0 \to I^ n\mathcal{F}_{n + 1}$ is an isomorphism for all $n \geq 0$. Thus it suffices to show that

$\bigoplus \nolimits _{n \geq 1} H^1(X, \mathcal{F}_1) \cdot f^{n + 1}$

is a graded $S = \bigoplus _{n \geq 0} A/(f) \cdot f^ n$-module satisfying the ascending chain condition. If $A$ is not Noetherian, then $H^1(X, \mathcal{F}_1)$ has finite length and the result holds. If $A$ is Noetherian, then $S$ is a Noetherian ring and the result holds as the module is finite over $S$ by the assumed finiteness of $H^1(X, \mathcal{F}_1)$. Some details omitted. $\square$

Lemma 52.3.5. Let $A$ be a ring. Let $f \in A$. Let $X$ be a scheme over $\mathop{\mathrm{Spec}}(A)$. Let

$\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$

be an inverse system of $\mathcal{O}_ X$-modules. Assume

1. either there is an $m \geq 1$ such that the image of $H^1(X, \mathcal{F}_ m) \to H^1(X, \mathcal{F}_1)$ is an $A$-module of finite length or $A$ is Noetherian and the intersection of the images of $H^1(X, \mathcal{F}_ m) \to H^1(X, \mathcal{F}_1)$ is a finite $A$-module,

2. the equivalent conditions of Lemma 52.3.1 hold.

Then the inverse system $M_ n = \Gamma (X, \mathcal{F}_ n)$ satisfies the Mittag-Leffler condition.

Proof. Set $I = (f)$. We will use the criterion of Lemma 52.2.2 involving the modules $H^1_ n$. For $m \geq n$ we have $I^ n\mathcal{F}_{m + 1} = \mathcal{F}_{m + 1 - n}$. Thus we see that

$H^1_ n = \bigcap \nolimits _{m \geq 1} \mathop{\mathrm{Im}}\left( H^1(X, \mathcal{F}_ m) \to H^1(X, \mathcal{F}_1) \right)$

is independent of $n$ and $\bigoplus H^1_ n = \bigoplus H^1_1 \cdot f^{n + 1}$. Thus we conclude exactly as in the proof of Lemma 52.3.4. $\square$

Lemma 52.3.6. Let $A$ be a ring and $f \in A$. Let $X$ be a scheme over $A$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume that $\mathcal{F}[f^ n] = \mathop{\mathrm{Ker}}(f^ n : \mathcal{F} \to \mathcal{F})$ stabilizes. Then

$R\Gamma (X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) = R\Gamma (X, \mathcal{F})^\wedge$

where the right hand side indicates the derived completion with respect to the ideal $(f) \subset A$. Let $H^ p$ be the $p$th cohomology group of this complex. Then there are short exact sequences

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/f^ n\mathcal{F}) \to H^ p \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/f^ n\mathcal{F}) \to 0$

and

$0 \to H^0(H^ p(X, \mathcal{F})^\wedge ) \to H^ p \to T_ f(H^{p + 1}(X, \mathcal{F})) \to 0$

where $T_ f(-)$ denote the $f$-adic Tate module as in More on Algebra, Example 15.93.5.

\begin{align*} R\Gamma (X, \mathcal{F})^\wedge & = R\mathop{\mathrm{lim}}\nolimits R\Gamma (X, \mathcal{F}) \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A) \\ & = R\mathop{\mathrm{lim}}\nolimits R\Gamma (X, \mathcal{F} \xrightarrow {f^ n} \mathcal{F}) \\ & = R\Gamma (X, R\mathop{\mathrm{lim}}\nolimits (\mathcal{F} \xrightarrow {f^ n} \mathcal{F})) \end{align*}

The first equality holds by More on Algebra, Lemma 15.91.18. The second by the projection formula, see Cohomology, Lemma 20.51.3. The third by Cohomology, Lemma 20.35.2. Note that by Derived Categories of Schemes, Lemma 36.3.2 we have $\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n \mathcal{F}$. Thus to finish the proof of the first statement of the lemma it suffices to show that the pro-objects $(f^ n : \mathcal{F} \to \mathcal{F})$ and $(\mathcal{F}/f^ n \mathcal{F})$ are isomorphic. There is clearly a map from the first inverse system to the second. Suppose that $\mathcal{F}[f^ c] = \mathcal{F}[f^{c + 1}] = \mathcal{F}[f^{c + 2}] = \ldots$. Then we can define an arrow of inverse systems in $D(\mathcal{O}_ X)$ in the other direction by the diagrams

$\xymatrix{ \mathcal{F}/\mathcal{F}[f^ c] \ar[r]_-{f^{n + c}} \ar[d]_{f^ c} & \mathcal{F} \ar[d]^1 \\ \mathcal{F} \ar[r]^{f^ n} & \mathcal{F} }$

Since the top horizontal arrow is injective the complex in the top row is quasi-isomorphic to $\mathcal{F}/f^{n + c}\mathcal{F}$. Some details omitted.

Since $R\Gamma (X, -)$ commutes with derived limits (Injectives, Lemma 19.13.6) we see that

$R\Gamma (X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) = R\Gamma (X, R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) = R\mathop{\mathrm{lim}}\nolimits R\Gamma (X, \mathcal{F}/f^ n\mathcal{F})$

(for first equality see first paragraph of proof). By More on Algebra, Remark 15.86.9 we obtain exact sequences

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/f^ n\mathcal{F}) \to H^ p(X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/I^ n\mathcal{F}) \to 0$

of $A$-modules. The second set of short exact sequences follow immediately from the discussion in More on Algebra, Example 15.93.5. $\square$

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